To evaluate the mesh current \(i_1\) in the given circuit, we will apply Kirchhoff's Voltage Law (KVL) to each mesh. The circuit consists of two meshes, and we will define the mesh currents as \(i_1\) for the left mesh and \(i_2\) for the right mesh.

Mesh 1 (Left Mesh):

• Apply KVL: \(-5 + 4i_1 + 2(i_1 - i_2) + 2 = 0\)
• Simplifying, we get: \(4i_1 + 2i_1 - 2i_2 - 3 = 0\)
• This can be further simplified to: \(6i_1 - 2i_2 = 3\)   &;(Equation 1)

Mesh 2 (Right Mesh):

• Apply KVL: \(-2 + 2(i_2 - i_1) + 5i_2 + 1 = 0\)
• Simplifying, we get: \(2i_2 - 2i_1 + 5i_2 - 1 = 0\)
• This can be further simplified to: \(7i_2 - 2i_1 = 1\)   &;(Equation 2)

Solve the Equations:

• Equations are: \(6i_1 - 2i_2 = 3\) (Equation 1)
• \(7i_2 - 2i_1 = 1\) (Equation 2)

Rearrange Equation 2: \(2i_1 = 7i_2 - 1 \Rightarrow i_1 = \frac{7i_2 - 1}{2}\)

• Substitute \(i_1\) from Equation 2 into Equation 1: \(6\left(\frac{7i_2 - 1}{2}\right) - 2i_2 = 3\)
• Simplify the equation: \(21i_2 - 3 - 2i_2 = 3\)
• \(19i_2 = 6\)
• Solving for \(i_2\): \(i_2 = \frac{6}{19}\)
• Substitute back to find \(i_1\): \(i_1 = \frac{7\left(\frac{6}{19}\right) - 1}{2}\)
• \(i_1 = \frac{42/19 - 1}{2} = \frac{23}{19}\)
• Simplified, we get: \(i_1 \approx 1.132 \, A\)

Conclusion:

The correct option is \(i_1 = 1.132 \, A\).